Question

If $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}\left( x\ne 3,-5 \right)$ , then x is equal to
A. -9, 7
B. 2, 3
C. 2, 1
D. None of the above

Answer 1

Hint: We need to find the value of the x in the equation $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}$ . We start to solve the given question by simplifying the given equation in the form of $a{{x}^{2}}+bx+c$ . Then, the value of x can be found out using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get the desired result.

Complete step by step solution:
We are given an equation and are asked to find the value of x in the equation. We will be solving the given question using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
The Quadratic equations are the polynomials with degree two. The quadratic equation will always have two roots. The roots may be real or imaginary.
The quadratic equation in standard form is given as follows,
$\Rightarrow f\left( x \right)=a{{x}^{2}}+bx+c$
Here,
a is the coefficient of ${{x}^{2}}$
b is the coefficient of $x$
c is the constant term
The roots of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ are found out using the formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
According to our question,
We need to simplify the equation
$\Rightarrow \dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}$
We know that the value of the expression $\dfrac{a}{b}-\dfrac{c}{d}$ can be written as $\dfrac{\left( a\times d \right)-\left( b\times c \right)}{\left( b\times d \right)}$
Following the same for the given equation, we get,
$\Rightarrow \dfrac{\left( x+5 \right)-\left( x-3 \right)}{\left( x+5 \right)\left( x-3 \right)}=\dfrac{1}{6}$
Simplifying the numerator on the left-hand side, we get,
$\Rightarrow \dfrac{x+5-x+3}{\left( x+5 \right)\left( x-3 \right)}=\dfrac{1}{6}$
$\Rightarrow \dfrac{8}{\left( x+5 \right)\left( x-3 \right)}=\dfrac{1}{6}$
Cross multiplying the above equation, we get,
$\Rightarrow 8\times 6=\left( x+5 \right)\left( x-3 \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( x+5 \right)\left( x-3 \right)=48$
Multiplying the terms inside the brackets, we get,
$\Rightarrow {{x}^{2}}-3x+5x-15=48$
Moving the number 48 to the other side of the equation, we get,
$\Rightarrow {{x}^{2}}+2x-15-48=0$
Simplifying the above equation, we get,
$\Rightarrow {{x}^{2}}+2x-63=0$
Comparing the quadratic equation ${{x}^{2}}+2x-63$ with the standard form $a{{x}^{2}}+bx+c$ , we get,
a = 1;
b = 2;
c = -63.
We need to find the roots of the quadratic equation ${{x}^{2}}+2x-63$
The roots of the quadratic equation ${{x}^{2}}+2x-63$ are found out using the formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of a, b, c in the above formula, we get,
$\Rightarrow x=\dfrac{-2\pm \sqrt{{{2}^{2}}-\left( 4\times 1\times -63 \right)}}{2\times 1}$
Simplifying the above equation, we get,
$\Rightarrow x=\dfrac{-2\pm \sqrt{4-\left( -252 \right)}}{2\times 1}$
Let us evaluate it further.
$\Rightarrow x=\dfrac{-2\pm \sqrt{4+252}}{2\times 1}$
$\Rightarrow x=\dfrac{-2\pm \sqrt{256}}{2\times 1}$
We know that the value of $\sqrt{256}$ is equal to 16. Substituting the same, we get,
$\Rightarrow x=\dfrac{-2\pm 16}{2}$
From the above,
$\Rightarrow x=\dfrac{-2+16}{2}\text{ or }x=\dfrac{-2-16}{2}$
Simplifying the above equations, we get,
$\Rightarrow x=\dfrac{14}{2}\text{ or }x=\dfrac{-18}{2}$
Canceling the common factors, we get,
$\therefore x=7\text{ and }x=-9$

So, the correct answer is “Option A”.

Note: The result of the given question can be cross-checked by substituting the value of x in the equation $\dfrac{1}{x-3}-\dfrac{1}{x+5}=\dfrac{1}{6}$ as follows,
The value of the LHS must be equal to RHS for the value of x equal to 7, -9.
For the value of x = 7,
LHS:
$\Rightarrow \dfrac{1}{x-3}-\dfrac{1}{x+5}$
Substituting the value of x = 7, we get,
$\Rightarrow \dfrac{1}{7-3}-\dfrac{1}{7+5}$
Simplifying the above expression, we get,
$\Rightarrow \dfrac{1}{4}-\dfrac{1}{12}$
Taking the LCM, we get,
$\Rightarrow \dfrac{3-1}{12}$
$\Rightarrow \dfrac{2}{12}$
Canceling the common factors, we get,
$\Rightarrow \dfrac{1}{6}$
RHS:
$\Rightarrow \dfrac{1}{6}$
For the value of x = -9,
LHS:
$\Rightarrow \dfrac{1}{x-3}-\dfrac{1}{x+5}$
Substituting the value of x = -9, we get,
$\Rightarrow \dfrac{1}{-9-3}-\dfrac{1}{-9+5}$
Simplifying the above expression, we get,
$\Rightarrow -\dfrac{1}{12}-\dfrac{1}{-4}$
$\Rightarrow -\dfrac{1}{12}+\dfrac{1}{4}$
Taking the LCM, we get,
$\Rightarrow \dfrac{3-1}{12}$
$\Rightarrow \dfrac{2}{12}$
Canceling the common factors, we get,
$\Rightarrow \dfrac{1}{6}$
RHS:
$\Rightarrow \dfrac{1}{6}$
LHS = RHS in both cases. The result attained is correct.